Problem: What is the value of the following logarithm? $\log_{3} \left(\dfrac{1}{27}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $3^{y} = \dfrac{1}{27}$ In this case, $3^{-3} = \dfrac{1}{27}$, so $\log_{3} \left(\dfrac{1}{27}\right) = -3$.